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- Jan 26th 2013, 12:09 PMmulferinequality
- Jan 26th 2013, 10:35 PMearthboyRe: inequality
Small advice @ mulfer:please learn LATEX

We have to prove:

$\displaystyle a^2+b^2+c^2 \geq a+b+c$

It is equivalent to proving:

$\displaystyle a^2+b^2+c^2-a-b-c \geq 0$

$\displaystyle \implies a(a-1)+b(b-1)c(c-1) \geq 0 $ So we are done if we are able to prove this.

now,

if $\displaystyle x$ is a positive real number,

if$\displaystyle x>1$ then $\displaystyle x(x-1)>(x-1)$ (obvious)

if $\displaystyle x=1 $then $\displaystyle x(x-1)=(x-1) $(obvious)

if $\displaystyle 0<x<1$ then $\displaystyle (x-1)<0 $ so, $\displaystyle x<1$ $\displaystyle \implies x(x-1)>x-1$ (multiplying both sides by a negative number i.e $\displaystyle (x-1)$ changes the sign)

so for all case, $\displaystyle x(x-1)\geq x-1$

similarly,

$\displaystyle a(a-1) \geq a-1$

$\displaystyle b(b-1) \geq b-1$

$\displaystyle c(c-1) \geq c-1$

summing these we get:

$\displaystyle a(a-1)+b(b-1)c(c-1) \geq (a-1)+(b-1)+(c-1) = (a+b+c)-3$

so we will be done if we prove $\displaystyle (a+b+c)-3 \geq 0$ or $\displaystyle a+b+c \geq 3$

by the AM-GM inequality,

$\displaystyle (a+b+c)\geq 3(abc)^{1/3}=3$ # as $\displaystyle abc=1$

proved (Rock) - Jan 27th 2013, 04:14 AMmulferRe: inequality
thanks for replying. I'd be glad to learn the latex soon.

please, check my solution:

Attachment 26722 - Jan 27th 2013, 05:32 AMearthboyRe: inequality